Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 41}{x - 7} = \dfrac{-3x + 111}{x - 7}$
Multiply both sides by $x - 7$ $ \dfrac{x^2 + 41}{x - 7} (x - 7) = \dfrac{-3x + 111}{x - 7} (x - 7)$ $ x^2 + 41 = -3x + 111$ Subtract $-3x + 111$ from both sides: $ x^2 + 41 - (-3x + 111) = -3x + 111 - (-3x + 111)$ $ x^2 + 41 + 3x - 111 = 0$ $ x^2 - 70 + 3x = 0$ Factor the expression: $ (x - 7)(x + 10) = 0$ Therefore $x = 7$ or $x = -10$ At $x = 7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 7$, it is an extraneous solution.